PAT练习（4）-1007. Maximum Subsequence Sum (25)

题目地址：https://www.nowcoder.com/pat/1/problem/3997

题目描述

Given a sequence of K integers { N1
, N2
, ..., NK
}.  A continuous subsequence is defined to be { Ni
, Ni+1
, ..., Nj
} where 1 <= i <= j <= K.  The Maximum Subsequence
is the continuous subsequence which has the largest sum of its
elements.  For example, given sequence { -2, 11, -4, 13, -5, -2 }, its
maximum subsequence is { 11, -4, 13 } with the largest sum being 20.

Now you are supposed to find the largest sum, together with the first
and the last numbers of the maximum subsequence.

输入描述:

Each input file contains one test case. Each case occupies two lines.  The first line contains a positive integer K (<= 10000).  The second line contains K numbers, separated by a space.

输出描述:

For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line.  In case that the maximum subsequence is not unique, output the one with the smallest indices i and j (as shown by the sample case).  If all the K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.

输入例子:

10
-10 1 2 3 4 -5 -23 3 7 -21

输出例子:

10 1 4

思路

当累加到第i个数字的时候：
如果前序子列的和为负数，那么前面的子列对于增大子列和是不利的，需要去掉，所以sum设为第i个数字；
如果前序子列的和为0或正数，那么前面的子列对于增大子列和是不利的，需要保留，所以sum设为第i个数字+前面的子列之和。

代码

#include <iostream>
#include <limits>
using namespace std;

int main()
{
int n;
cin >> n;

bool nagative = true;

int max = 0, sum = 0;
int start = 0, end = 0;
int tstr = 0, tend = 0;

int firstnum, lastnum;

for (int i = 0; i < n; i++)
{
if (i == 0)
{
cin >> max;
sum = max;
start = max;
end = max;
tstr = max;
tend = max;

firstnum = max;
lastnum = max;
if (max >= 0)
nagative = false;

continue;
}

int tmp;
cin >> tmp;

if (tmp >= 0)
nagative = false;
lastnum = tmp;

if (sum < 0)
{
sum = tmp;
tstr = tmp;
tend = tmp;
if (sum > max)
{
max = sum;
start = tstr;
end = tend;
}
continue;
}

if (sum+tmp>=0)
{
sum += tmp;
tend = tmp;
if (sum > max)
{
max = sum;
start = tstr;
end = tend;
}
//else if (sum == max && (tend - tstr > end - start))
//{
//	max = sum;
//	start = tstr;
//	end = tend;
//}
}
else
{
sum = tmp;
tstr = tmp;
tend = tmp;
}
}

if (nagative)
cout << 0 << " " << firstnum << " " << lastnum;
else
cout << max << " " << start << " " << end;

return 0;
}